Fragmentation Network layer

Question

I got answer 12 is it Right or wrong

Comments

I don't answer that why I did ask this question for verification.

I got 12 because in first router 4 fragments and in second each divided into 3 so

4*3

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actually i was not cleared with your qsn, if this is a case it will be '11'...first router 4 fragments...out of these 3 will be divided into 3 and one will be divided into 2 fragments in B....

3*3+1*2= 11

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I do the same
Can you post your solution

Answer 1

Answer should be 11.  
MTU of Router A = 1200B, it can carry 1180 Bytes of Data + 20 Bytes IP Header
But 1180 is not divisible by 8. hence MAX packet will carry 1176B + 20B (IP header)
MTU of Router B = 512B, it can carry 492 Bytes of Data + 20 Bytes IP Header, but same as said above
492 is not divisible by 8, hence Max packet will carry 488B + 20B (IP Header) 

IP Packet is of 4408 B, Means there is 4388 Bytes of Data + 20 Bytes IP. 
At router A there will be 3packets of size 1176 Bytes each and 4th packet will be of size 880, in addition to this each packet will carry 20 Bytes IP header. 
  
At router B, 488B Data + 20B Header. each packet of 1176 B will be divided into three parts and 20B of IP header will be added to each. 
There are three packets of 1180 B,and each is divided into 3 fragements. 3*3 = 9 
Fourth packet which was carrying 880B + 20B IP header will be divided into two parts at router B. 
So total fragemtens will be 11.

PS: MTU header is not considered when we talk about MTU size.

Comments

last packet size need to be in the multiple of 8. this can be decided for each packet which has MF=0

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@manu00x I mean to say that after creating 3 packets having data size as 1176 at A, Now the data left to be send is 4388-3*1176 = 860 Bytes not 880 Bytes. In your solution you have written "4th packet will be of size 880", and 860 not div by 8 so 864 Bytes be the size of 4th packet.

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@Manu Not that it matters but we can take last packet as 864 right? 864 is the closest to be divisible by 8.